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Simple Circuit

Discussion in 'Electronic Basics' started by [email protected], Apr 13, 2005.

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  1. Guest

    Hello,

    This is my first time posting to USENET. I was wondering if anyone
    could help me figure out if I have the correct circuit for what I am
    trying to do. I have followed other peoples schematics before, and this
    is the first time that I have tried to make something for myself.

    You can see the schematics that I drew up here:
    http://img68.echo.cx/my.php?image=schem1wn.png

    I want to run 4 things off of one power supply. The 15V 4.5A is an old
    laptop power adapter. The two fans run at 12V and .01A-.02A. The LED
    runs from 3.5V-4V. The peltier will run upto 15V and 15A. I did'nt have
    any images for the fans and peltier so I just used some random ones.

    What I don't understand is does a device draw as many amps as it needs?
    like the fans and the LED or does that need to be limited some how?

    Thanks
     
  2. Chris

    Chris Guest

    Hi. You might want to replace the LM317T with a simple
    current-limiting resistor. Diodes are current-driven devices. Your
    white LED may have 3.5V across it if you put 20 mA through it, and it
    will only have 3.7V across it if it has a lethal 50 mA. By the time it
    gets to 3.9V, you might just have a smoking blob instead of an LED.
    Just use a 560 ohm, 1/2 watt resistor in series with the LED, and you
    should be fine. That will give you about 20mA through the LED.

    Using an LM7812 as a series regulator for the fans is a good idea.
    Your schematic doesn't show a connect between the GND pin of the 7812
    and GND -- I'd guess that's just an oversight. You should place a
    small (10uF) cap from input to GND, and from output to GND, to avoid
    oscillations which might cause problems.

    The part that's bothersome, though, is the peltier cooler. You say
    you've got a cooler that needs 15V at up to 15A, but your power supply
    is only capable of 15V at 4.5A. If that's not a typo, you're in
    trouble. Choose a power supply that will provide the voltage you want,
    but make sure it can provide enough current.

    If you've got a voltage source, it's rated to provide a certain amount
    of current. That means you can't exceed that amount, or something
    won't work right. It might be automatic shutdown, current limiting
    (the power supply drops down to a lower voltage to keep current below
    maximum), or the power supply might be damaged or even smoke. A good
    power supply will be able to provide a steady DC voltage over a wide
    range of current, as long as it's less than the maximum. As an extra
    note, you should know that some switching power supplies require a
    minimum load current, too. It's best to ask before specifying.

    Good luck
    Chris


    Good luck
    Chris
     
  3. Andrew Holme

    Andrew Holme Guest

    With the fans, you're OK - they'll "draw as many amps as they need," but you
    need a current-limiting resistor in series with LED. You would be
    better-off connecting the LED, in series with a resistor, across the 12V or
    15V supply. Forget about VREG1 altogether. Calculate the LED current using
    Ohm's law:

    I = (Vs - Vf) / Rs

    Vs = Supply voltage
    Vf = Approximate forward voltage drop across LED (3.5V ish)
    Rs = Series resistor

    You need decoupling capacitors (try 100nF) on the input and output of the
    other (remaining) voltage regulator.
     
  4. One day got dressed and committed to text
    Yup, things that are designated as 12v 15v etc will only take what current
    they need.
    An LED needs to have the current limited, your schematic would likely
    'terminate' the LED. While LED's have a forward voltage you have to deduct
    that from the supply voltage and then calculate a resistance that will limit
    the current. Say you have taken 3.3v as the LED forward voltage, 15v - 3.3v
    = 11.7v .
    So to limit the current to 20mA a resistor of 11.7v / .02mA is required,
    which would be 585ohms (use 680ohms). No need for the voltreg here just the
    resistor. Note that you can use a lower current for very little loss of LED
    output, if power is an issue 10mA or less will still give a good output
    (particularly with high intensity LED')s.
     
  5. Guest

  6. Lord Garth

    Lord Garth Guest

    When digital circuits change states, there is a very brief time when both
    output devices are conducting. That creates a spike on the power rails
    which you can easily see on a scope. The caps absorb much of this by
    providing a local source of power.
     
  7. In this situation the caps restrain the regulator from oscillating.

    Once more, switching on the Peltier will switch off or blow your power
    supply.

    petrus bitbyter
     
  8. Andrew Holme

    Andrew Holme Guest

    Since it contains a feedback loop, the regulator has the potential to
    oscillate - depending on the load impedance. It is standard practice
    to decouple the regulator inputs and outputs. Look at the typical
    application circuit on the datasheet. The capacitors must be mounted
    close to the regulator to be effective.
     
  9. Guest

    How do I keep the peltier from trying pull more amps than the
    powersupply can handle with out droping the voltage?

    Thanks
     
  10. Lord Garth

    Lord Garth Guest

    You don't! You make sure your power supply has the necessary head room and
    let the peltier take what it needs.
     
  11. If you built a switching buck regulator, you could lower the supply
    voltage to the peltier so it would draw less current (and move less
    heat). The current demand is roughly proportional to the supply
    voltage. So, for instance, if you lowered the 15 volt supply to 10
    volts, the peltier current would fall from 15 amps to 10 for a total
    power to the peltier of 10 volts * 10 amperes = 100 watts.

    Assuming the efficiency of the buck regulator was 90%, its input
    current would be (100 watts * 100/90%)/ 15 volts = 7.4 amps.
     
  12. Guest

    Are those the maximum ratings for the Peltier? In practice, their
    efficiency drops very low as you approach the maximum voltage and
    current, as they will internally generate more heat than they are
    pumping.

    Best to run it at roughly half of its voltage and current rating, or
    less.

    Mark
     
  13. Guest

    Would I have to build a buck regulator like John Popelish described in
    order to limit the current.
     
  14. redbelly

    redbelly Guest

    A buck regulator is just one way to do this. You could also try getting
    another power supply.

    Mark
     
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